1. ## Mean Deviation.

I am doing some work with moment generating functions and have been given the question ''calculate the mean deviation'', this is defined to be E(lX-E(X)l).

My m.g.f is (1-t/lambda)^-1 and was generated from f(x)=(lambda)e^-(lambda)x.

I have no idea how to find MD from a m.g.f.

cheers

2. Originally Posted by ardam
I am doing some work with moment generating functions and have been given the question ''calculate the mean deviation'', this is defined to be E(lX-E(X)l).

My m.g.f is (1-t/lambda)^-1 and was generated from f(x)=(lambda)e^-(lambda)x.

I have no idea how to find MD from a m.g.f.

cheers
Here is a blunt approach:

Get $\displaystyle E(X) = \frac{1}{\lambda}$ from the mgf. Now use the pdf to calculate $\displaystyle E\left(\left |X - \frac{1}{\lambda} \right|\right)$.

3. Originally Posted by mr fantastic
Here is a blunt approach:

Get $\displaystyle E(X) = \frac{1}{\lambda}$ from the mgf. Now use the pdf to calculate $\displaystyle E\left(\left |X - \frac{1}{\lambda} \right|\right)$.
Got $\displaystyle E(X) = \frac{1}{\lambda}$ by differentiating m.g.f once then setting t=0.

So do i manipulate the expectation of a random variable formula to look like:
$\displaystyle E\left(\left |X - \frac{1}{\lambda} \right|\right)$ = intergral |X - (1\lambda)| (lambda)e^-(lambda)x. dx?

4. Originally Posted by ardam
Got $\displaystyle E(X) = \frac{1}{\lambda}$ by differentiating m.g.f once then setting t=0.

So do i manipulate the expectation of a random variable formula to look like:
$\displaystyle E\left(\left |X - \frac{1}{\lambda} \right|\right)$ = intergral |X - (1\lambda)| (lambda)e^-(lambda)x. dx?
Yes. And note that $\displaystyle \left |X - \frac{1}{\lambda} \right| = X - \frac{1}{\lambda}$ when $\displaystyle X > \frac{1}{\lambda}$ and $\displaystyle \left |X - \frac{1}{\lambda} \right| = \frac{1}{\lambda} - X$ when $\displaystyle X < \frac{1}{\lambda}$.